Re: Solve output depends on previous attempt with bad syntax
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- Subject: [mg132512] Re: Solve output depends on previous attempt with bad syntax
- From: Alain.Cochard at unistra.fr
- Date: Fri, 4 Apr 2014 03:57:18 -0400 (EDT)
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- Reply-to: alain.cochard at unistra.fr
Hi Kevin, and thank you very much. Kevin writes on Thu 3 Apr 2014 06:29: > You could have anticipated this by noting two things: > > (1) The symbol y1 is now black instead of blue, which means that it = has > a value (b+a x1) in this case; Hum, this rather looks like an aposteriori hint! At any rate, I use the text based interface, which has no color variations. > (2) The use of the "=" itself is the reason. > > The second execution was looking for a solution to > > a x1 + b == a x1 +b, a x2 + b == a x2 + b > > Which is why you got the null answer, since a,b can be anything. > > To prevent this and other problems with previously used variables, I= > would generally include > > Clear[a,b,x1,x2,y1,y2] > > in the same cell as the Solve. OK, I see the reason now. I dare say this is a some sort of a misconception... Regards, Alain > Kevin > > > On 4/3/2014 2:15 AM, Alain Cochard wrote: > > First, the correct solution of a linear system of 2 equations with= 2 > > unknowns, for reference. > > > > Mathematica 9.0 for Linux x86 (64-bit) > > Copyright 1988-2013 Wolfram Research, Inc. > > > > In[1]:= Solve[{y1==a x1 +b, y2==a x2 +b},{a,b}] > > > > -y1 + y2 x2 y1 - x1 y2 > > Out[1]= {{a -> -(--------), b -> -(-------------)}} > > x1 - x2 x1 - x2 > > > > Now, starting a new Mathematica session, assume I make a mistake, = using '=' > > instead of '==', I get an error message -- so far, so good: > > > > Mathematica 9.0 for Linux x86 (64-bit) > > Copyright 1988-2013 Wolfram Research, Inc. > > > > In[1]:= Solve[{y1=a x1 +b, y2=a x2 +b},{a,b}] > > > > Solve::naqs: b + a x1 && b + a x2 is not a quantified system > > of equations and inequalities. > > > > Out[1]= Solve[{b + a x1, b + a x2}, {a, b}] > > > > Realizing my mistake, I retry with the proper syntax: > > > > In[2]:= Solve[{y1==a x1 +b, y2==a x2 +b},{a,b}] > > > > Out[2]= {{}} > > > > which is obviously not the correct result. > > > > Is there a rationale here, i.e., could I have anticipated this out= put=3F > > > > Thank you, > > Alain > > -- EOST (=C9cole et Observatoire des Sciences de la Terre) IPG (Institut de Physique du Globe) | alain.cochard at unistra.fr 5 rue Ren=E9 Descartes [bureau 106] | Phone: +33 (0)3 68 85 50 44 F-67084 Strasbourg Cedex, France | Fax: +33 (0)3 68 85 01 25 =